Masses `M_(1),M_(2) and M_(3)` are connected by string of negligible mass which pass over massless and frictionless pulleys `P_(1) and P_(2)` as shown in figure 7.15. The masses move such that the string between `P_(1) and P_(2)` is parallel to the incline and the portion of the string between `P_(2) and M_(3)` is horizontal. The masses `M_(2) and M_(3)` are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfces is 0.25. The inclined plane makes an angle of `37^(@)` with the horizontal. If the mass `M_(1)`moves downwards with a uniform velocity, find (i) the mass of `M_(1)`, (ii) the tension in the horizontal portion. `(g=9.8 ms^(-2) , sin 37^(@)=(3)/(5))`

Answer

(i) 4.2 (ii) 10

Explanation

(i) The block `M_(1)` is moving with constant speed
`m_(1) g - T = 0 implies T - M_(1) g`

The Block `m_(1)` is also moving with constant speed, we get
`T = T' + M_(2) g sin theta f = T' + M_(2) g sin theta + mu N_(2)` `[:. f_(2) = muN_(2) = mu M_(2) g cos theta]`
(ii) `:. T = T' + M_(2) g sin theta + mu M_(2) g cos theta`
For `M_(3)` we get
`T' - f_(3) = M_(3) xx 0` `implies` `T' = f_(3) = mu M_(3) g`
Putting the value of T and T' from (i) and (iii) in (ii) we get
`M_(1) g = mu N_(3) g + M_(2) g sin theta + mu M_(2) g cos theta`
After substituting the values, we get, `M_(1) = 4.2 kg`
The tension in the horizontal string will be
`T' = mu M_(3) g = 0.25 xx 4 xx 10 implies T' = 10 N`